I was given the following exercise.
Let $X$ be a random variable that takes non-negative natural number
values such that $E[X]=1,E[X^2]=2,E[X^3]=5$. Find the minimum value of
$P(X=0)$ using the taylor expansion of the probability generating function
at $z=1$.
I know the method not using the generating function stated in this question.
My attempt:
Let $G(z)=E[z^X]$ be the probability generating function of $X$. Then by definition,
$G(z) = P(X=0)+P(X=1)z+P(X=2)z^2+\cdots $
Also, since $E[\frac{X(X-1)\cdots (X-n+1)}{n!}]=\frac{G^{(n)}(1)}{n!}$ according to Wikipedia, we have
$\displaystyle G(z) = \sum_{n=0}^{\infty}E\left[\frac{X(X-1)\cdots (X-n+1)}{n!}\right](z-1)^n$
Substituting $z=0$ yields
$\displaystyle P(X=0) = \sum_{n=0}^{\infty}E\left[\frac{X(X-1)\cdots (X-n+1)}{n!}\right](-1)^n$
Note that for $n=0,1,2,3$, the coefficient can be calculated as follows:
$\begin{align} E[1] &= 1 \\ E[X]&= 1 \\ E[X(X-1)/2] &= E[X^2]/2 -E[X]/2 = 1-1/2 =1/2 \\ E[X(X-1)(X-2)/3!] &= E[X^3]/6 -E[X^2]/2 +E[X]/3 = 5/6-2/2+1/3 \\ &= 1/6 \end{align}$
Therefore,
$\displaystyle \begin{align} P(X=0) &= \sum_{n=0}^{\infty}E\left[\frac{X(X-1)\cdots (X-n+1)}{n!}\right](-1)^n \\ &=1-1+\frac{1}{2} -\frac{1}{6} +\sum_{n=4}^{\infty}E\left[\frac{X(X-1)\cdots (X-n+1)}{n!}\right](-1)^n \\ &= \frac{1}{3} + \sum_{n=4}^{\infty}E\left[\frac{X(X-1)\cdots (X-n+1)}{n!}\right](-1)^n\end{align}$
According to the question linked above, $1/3$ is the minimum, so I think we need to prove that the sum is non-negative to complete the proof. However, I was unable to do so.
Am I on the right path? If not, what is the correct one? If yes, how can I finish it?
Best Answer
We have absolute convergence on the (complex) unit ball around $0$. Using the Taylor expansion of order three with an explicit form of the rest we have for some $c=c(z)\in(z,1)$: $$ \begin{aligned} G(z) &= \frac 1{0!}G(1) + \frac 1{1!}G'(1)(z-1) \\ &\qquad\qquad + \frac 1{2!}G''(1)(z-1)^2 + \frac 1{3!}G'''(1)(z-1)^3 \\ &\qquad\qquad\qquad\qquad + \frac 1{4!}G^{(IV)}(c)(z-1)^4 \ . \\ &\qquad\text{Taking $z=0$...} \\ G(0) &= \frac 1{0!}E[1] + \frac 1{1!}E[X](0-1) \\ &\qquad\qquad + \frac 1{2!}E[X(X-1)](0-1)^2 + \frac 1{3!}E[X(X-1)(X-2)](0-1)^3 \\ &\qquad\qquad\qquad\qquad + \frac 1{4!}E[\underbrace{X(X-1)(X-2)(X-3)\cdot c^{X-4}}_{\ge 0}](0-1)^4 \\ &\ge 1 -\frac 1{1!}\cdot 1+\frac 1{2!}(2-1)-\frac 1{3!}(5-3\cdot 2+2\cdot 1) \\ &=1-1+\frac 12-\frac 16=\frac 13 \ . \end{aligned} $$ $\square$
Note: The equality is satisfied if and only if for some (but then also for any) $c\in(0,1)$ we have $X(X-1)(X-2)(X-3)c^{X-4}=0$. This means that $X$ takes only the values $0,1,2,3$.
Note: The above tries to follow as close as possible the path from the OP. But there are some details related to the application of Taylor's Formula, if we insist to go this way. (Usually we want $G$ to be differentiable in the suitable class on an interval around the point we develop. This can be arranged as follows. Use some $a\in (0,1)$ instead of one first, then use the continuity of the $G$-derivatives to replace $a$ by $1^-$, and use their continuity in the compact ball of absolute convergence of the series for $G(z)$.)